Question

If $\frac{x^2+5x+1}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)(x+2)}+\frac{C}{(x+1)(x+2)(x+3)}$, then C=

• A. -5
• B. 8
• C. 0
• D. -6

Answer 1

Answer: (A)

-5

According to the question.......................

$\begin{array}{c}\frac{x^2+5x+1}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)(x+2)}+\frac{C}{(x+1)(x+2)(x+3)}\\ =\frac{A(x+2)(x+3)+B(x+3)+c}{(x+1)(x+2)(x+3)}\\ =\frac{A(x^2+5x+6)+Bx+3B+c}{(x+1)(x+2)(x+3)}\\ \frac{x^2+5x+1}{(x+1)(x+2)(x+3)}=\frac{A{x}^2+(5A+B)x+6A+3B+c}{(x+1)(x+2)(x+3)}\\ x^2+5x+1=A{x}^2+(5A+B)x+6A+3B+c\\ Getcofficientcompare:\\ x^2:1=A\\ x:5=5A+B\Rightarrow 5=5\times 1+B,B=0\\ cofficientterm:\\ 1=6A+3B+C------\left(i\right)\\ 1=6+C\\ C=-5\\ InwhichthevalueofA,B,Cis,\\ A=1,\\ B=0,\\ C=-5\\ So,thatthecorrectoptionisA.\end{array}$